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rewrite (5) and add a Remark on (5).
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JGWang
JGWang
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Here, use your symbols, a proof will be given under following condition (1): \begin{equation*} \lim_{m\to\infty}\mathsf{P}\Big(\sup_{0\le t\le 1}\Big| \frac{\alpha(mt)}{m} \Big| \ge\delta \Big)=0,\qquad \forall \delta>0. \tag{1} \end{equation*} Let \begin{align*} Y_m(t) &=\frac{1}{\sqrt{m}}X(mt), \tag{2} \\ Z_m(t) &=\frac{1}{\sqrt{m}}X\big(m(t+\tfrac{\alpha(mt)}{m})\big) =Y_m\big(t+\tfrac{\alpha(mt)}{m}\big), \tag{3}\\ U_m(t) &=Z_m(t)-Y_m(t)=Y_m\big(t+\tfrac{\alpha(mt)}{m}\big) - Y_m(t). \tag{4} \end{align*}

In the following, the concept of modulus of continuity of an arbitrary function $x(.)$ on $[0,1]$ also be used(cf. P. Billingsley, Convergence of Probability Measures, 2ed. $\S$7, pp 80-81). Let \begin{equation*} w(x,\delta)=\sup_{|s-t|\le \delta}|x(s)-x(t)|, \qquad 0<\delta \le 1. \end{equation*} For fixed $\delta>0$, since \begin{equation*} |w(x,\delta)-w(y,\delta)|\le 2\sup\limits_{0\le t\le 1}|x(t)-y(t)|, \end{equation*} $w(x,\delta)$ is continuous in $x$, using this and $Y_m\stackrel{d}{\longrightarrow}B$(i.e., the distributions of $Y_m$ converge to to the distribution of BM $B$ weakly on $C[0,1]$), get \begin{gather*} \lim_{m\to\infty}w(Y_m,\delta)\stackrel{d}{=} w(B,\delta).\\ \lim_{\delta\to0}\varlimsup_{m\to\infty} \mathsf{P}(w(Y_m,\delta)>\epsilon)=0, \quad \forall \epsilon>0.\tag{5} \end{gather*}\begin{gather*} \lim_{m\to\infty}w(Y_m,\delta)\stackrel{d}{=} w(B,\delta), \qquad \forall \delta>0,\\ \varlimsup_{m\to\infty} \mathsf{P}(w(Y_m,\delta)\ge \epsilon)\le \mathsf{P}(w(B,\delta) \ge \epsilon),\quad \forall \epsilon,\delta>0, \\ \lim_{\delta\to0}\varlimsup_{m\to\infty} \mathsf{P}(w(Y_m,\delta) \ge \epsilon)=0, \quad \forall \epsilon>0.\tag{5} \end{gather*} Next, the following fact will be proved, \begin{equation*} \lim_{m\to\infty}\mathsf{P}(\sup_{0\le t\le 1}|U_m(t)|>\epsilon)=0, \quad \forall \epsilon>0. \tag{6} \end{equation*} Denote \begin{equation*} A_m(\delta)=\Big\{\omega: \sup_{0\le t\le 1}\Big| \frac{\alpha(mt)}{m} \Big| \ge \delta \Big\}, \end{equation*} then from (1) \begin{gather*} \lim_{m\to\infty}\mathsf{P}(A_m(\delta))=0, \qquad \forall \delta>0,\\ \lim_{m\to\infty}\mathsf{P}([\sup_t|U_m(t)|]I_{A_m(\delta)}>\epsilon)=0, \qquad \forall \delta, \epsilon>0. \tag{7} \end{gather*} On the other hand, \begin{gather*} |U_m(t)| I_{A_m^c(\delta)}\le w(Y_m,\delta), \qquad \forall t\in[0,1],\\ [\sup_{0\le t\le 1}|U_m(t)|]I_{A_m^c(\delta)}\le w(Y_m,\delta). \tag{8} \end{gather*} From (8),(5), get \begin{equation*} \lim_{\delta\to0}\varlimsup_{m\to\infty} \mathsf{P}([\sup_{0\le t\le 1}|U_m(t)|]I_{A_m^c(\delta)}>\epsilon)=0, \quad \forall \epsilon>0.\tag{9} \end{equation*} Hence, from (7) and (9), (6) is true. Furthermore, $Z_m\stackrel{d}{\longrightarrow}B$ comes from $Y_m\stackrel{d}{\longrightarrow}B$ and (4),(6).

Remark on (5): Generally, for the sequence of random variables $V_m=w(Y_m,\delta), m\ge1$, if $V_m\stackrel{d}{\longrightarrow} V(=w(B,\delta))$, we could get only \begin{equation*} \mathsf{P}(V>\epsilon)\le \varliminf_{m\to\infty} \mathsf{P}(V_m>\epsilon) \le \varlimsup_{m\to\infty} \mathsf{P}(V_m\ge\epsilon)\le \mathsf{P}(V\ge\epsilon), \quad \forall \epsilon>0. \end{equation*}

Here, use your symbols, a proof will be given under following condition (1): \begin{equation*} \lim_{m\to\infty}\mathsf{P}\Big(\sup_{0\le t\le 1}\Big| \frac{\alpha(mt)}{m} \Big| \ge\delta \Big)=0,\qquad \forall \delta>0. \tag{1} \end{equation*} Let \begin{align*} Y_m(t) &=\frac{1}{\sqrt{m}}X(mt), \tag{2} \\ Z_m(t) &=\frac{1}{\sqrt{m}}X\big(m(t+\tfrac{\alpha(mt)}{m})\big) =Y_m\big(t+\tfrac{\alpha(mt)}{m}\big), \tag{3}\\ U_m(t) &=Z_m(t)-Y_m(t)=Y_m\big(t+\tfrac{\alpha(mt)}{m}\big) - Y_m(t). \tag{4} \end{align*}

In the following, the concept of modulus of continuity of an arbitrary function $x(.)$ on $[0,1]$ also be used(cf. P. Billingsley, Convergence of Probability Measures, 2ed. $\S$7, pp 80-81). Let \begin{equation*} w(x,\delta)=\sup_{|s-t|\le \delta}|x(s)-x(t)|, \qquad 0<\delta \le 1. \end{equation*} For fixed $\delta>0$, since \begin{equation*} |w(x,\delta)-w(y,\delta)|\le 2\sup\limits_{0\le t\le 1}|x(t)-y(t)|, \end{equation*} $w(x,\delta)$ is continuous in $x$, using this and $Y_m\stackrel{d}{\longrightarrow}B$(i.e., the distributions of $Y_m$ converge to to the distribution of BM $B$ weakly on $C[0,1]$), get \begin{gather*} \lim_{m\to\infty}w(Y_m,\delta)\stackrel{d}{=} w(B,\delta).\\ \lim_{\delta\to0}\varlimsup_{m\to\infty} \mathsf{P}(w(Y_m,\delta)>\epsilon)=0, \quad \forall \epsilon>0.\tag{5} \end{gather*} Next, the following fact will be proved, \begin{equation*} \lim_{m\to\infty}\mathsf{P}(\sup_{0\le t\le 1}|U_m(t)|>\epsilon)=0, \quad \forall \epsilon>0. \tag{6} \end{equation*} Denote \begin{equation*} A_m(\delta)=\Big\{\omega: \sup_{0\le t\le 1}\Big| \frac{\alpha(mt)}{m} \Big| \ge \delta \Big\}, \end{equation*} then from (1) \begin{gather*} \lim_{m\to\infty}\mathsf{P}(A_m(\delta))=0, \qquad \forall \delta>0,\\ \lim_{m\to\infty}\mathsf{P}([\sup_t|U_m(t)|]I_{A_m(\delta)}>\epsilon)=0, \qquad \forall \delta, \epsilon>0. \tag{7} \end{gather*} On the other hand, \begin{gather*} |U_m(t)| I_{A_m^c(\delta)}\le w(Y_m,\delta), \qquad \forall t\in[0,1],\\ [\sup_{0\le t\le 1}|U_m(t)|]I_{A_m^c(\delta)}\le w(Y_m,\delta). \tag{8} \end{gather*} From (8),(5), get \begin{equation*} \lim_{\delta\to0}\varlimsup_{m\to\infty} \mathsf{P}([\sup_{0\le t\le 1}|U_m(t)|]I_{A_m^c(\delta)}>\epsilon)=0, \quad \forall \epsilon>0.\tag{9} \end{equation*} Hence, from (7) and (9), (6) is true. Furthermore, $Z_m\stackrel{d}{\longrightarrow}B$ comes from $Y_m\stackrel{d}{\longrightarrow}B$ and (4),(6).

Here, use your symbols, a proof will be given under following condition (1): \begin{equation*} \lim_{m\to\infty}\mathsf{P}\Big(\sup_{0\le t\le 1}\Big| \frac{\alpha(mt)}{m} \Big| \ge\delta \Big)=0,\qquad \forall \delta>0. \tag{1} \end{equation*} Let \begin{align*} Y_m(t) &=\frac{1}{\sqrt{m}}X(mt), \tag{2} \\ Z_m(t) &=\frac{1}{\sqrt{m}}X\big(m(t+\tfrac{\alpha(mt)}{m})\big) =Y_m\big(t+\tfrac{\alpha(mt)}{m}\big), \tag{3}\\ U_m(t) &=Z_m(t)-Y_m(t)=Y_m\big(t+\tfrac{\alpha(mt)}{m}\big) - Y_m(t). \tag{4} \end{align*}

In the following, the concept of modulus of continuity of an arbitrary function $x(.)$ on $[0,1]$ also be used(cf. P. Billingsley, Convergence of Probability Measures, 2ed. $\S$7, pp 80-81). Let \begin{equation*} w(x,\delta)=\sup_{|s-t|\le \delta}|x(s)-x(t)|, \qquad 0<\delta \le 1. \end{equation*} For fixed $\delta>0$, since \begin{equation*} |w(x,\delta)-w(y,\delta)|\le 2\sup\limits_{0\le t\le 1}|x(t)-y(t)|, \end{equation*} $w(x,\delta)$ is continuous in $x$, using this and $Y_m\stackrel{d}{\longrightarrow}B$(i.e., the distributions of $Y_m$ converge to to the distribution of BM $B$ weakly on $C[0,1]$), get \begin{gather*} \lim_{m\to\infty}w(Y_m,\delta)\stackrel{d}{=} w(B,\delta), \qquad \forall \delta>0,\\ \varlimsup_{m\to\infty} \mathsf{P}(w(Y_m,\delta)\ge \epsilon)\le \mathsf{P}(w(B,\delta) \ge \epsilon),\quad \forall \epsilon,\delta>0, \\ \lim_{\delta\to0}\varlimsup_{m\to\infty} \mathsf{P}(w(Y_m,\delta) \ge \epsilon)=0, \quad \forall \epsilon>0.\tag{5} \end{gather*} Next, the following fact will be proved, \begin{equation*} \lim_{m\to\infty}\mathsf{P}(\sup_{0\le t\le 1}|U_m(t)|>\epsilon)=0, \quad \forall \epsilon>0. \tag{6} \end{equation*} Denote \begin{equation*} A_m(\delta)=\Big\{\omega: \sup_{0\le t\le 1}\Big| \frac{\alpha(mt)}{m} \Big| \ge \delta \Big\}, \end{equation*} then from (1) \begin{gather*} \lim_{m\to\infty}\mathsf{P}(A_m(\delta))=0, \qquad \forall \delta>0,\\ \lim_{m\to\infty}\mathsf{P}([\sup_t|U_m(t)|]I_{A_m(\delta)}>\epsilon)=0, \qquad \forall \delta, \epsilon>0. \tag{7} \end{gather*} On the other hand, \begin{gather*} |U_m(t)| I_{A_m^c(\delta)}\le w(Y_m,\delta), \qquad \forall t\in[0,1],\\ [\sup_{0\le t\le 1}|U_m(t)|]I_{A_m^c(\delta)}\le w(Y_m,\delta). \tag{8} \end{gather*} From (8),(5), get \begin{equation*} \lim_{\delta\to0}\varlimsup_{m\to\infty} \mathsf{P}([\sup_{0\le t\le 1}|U_m(t)|]I_{A_m^c(\delta)}>\epsilon)=0, \quad \forall \epsilon>0.\tag{9} \end{equation*} Hence, from (7) and (9), (6) is true. Furthermore, $Z_m\stackrel{d}{\longrightarrow}B$ comes from $Y_m\stackrel{d}{\longrightarrow}B$ and (4),(6).

Remark on (5): Generally, for the sequence of random variables $V_m=w(Y_m,\delta), m\ge1$, if $V_m\stackrel{d}{\longrightarrow} V(=w(B,\delta))$, we could get only \begin{equation*} \mathsf{P}(V>\epsilon)\le \varliminf_{m\to\infty} \mathsf{P}(V_m>\epsilon) \le \varlimsup_{m\to\infty} \mathsf{P}(V_m\ge\epsilon)\le \mathsf{P}(V\ge\epsilon), \quad \forall \epsilon>0. \end{equation*}

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JGWang
JGWang
  • 724
  • 4
  • 7

Here, use your symbols, a proof will be given under following condition (1): \begin{equation*} \lim_{m\to\infty}\mathsf{P}\Big(\sup_{0\le t\le 1}\Big| \frac{\alpha(mt)}{m} \Big| \ge\delta \Big)=0,\qquad \forall \delta>0. \tag{1} \end{equation*} Let \begin{align*} Y_m(t) &=\frac{1}{\sqrt{m}}X(mt), \tag{2} \\ Z_m(t) &=\frac{1}{\sqrt{m}}X\big(m(t+\tfrac{\alpha(mt)}{m})\big) =Y_m\big(t+\tfrac{\alpha(mt)}{m}\big), \tag{3}\\ U_m(t) &=Z_m(t)-Y_m(t)=Y_m\big(t+\tfrac{\alpha(mt)}{m}\big) - Y_m(t). \tag{4} \end{align*}

In the following, the concept of modulus of continuity of an arbitrary function $x(.)$ on $[0,1]$ also be used(cf. P. Billingsley, Convergence of Probability Measures, 2ed. $\S$7, pp 80-81). Let \begin{equation*} w(x,\delta)=\sup_{|s-t|\le \delta}|x(s)-x(t)|, \qquad 0<\delta \le 1. \end{equation*} For fixed $\delta>0$, since \begin{equation*} |w(x,\delta)-w(y,\delta)|\le 2\sup\limits_{0\le t\le 1}|x(t)-y(t)|, \end{equation*} $w(x,\delta)$ is continuous in $x$, using this and $Y_m\stackrel{d}{\longrightarrow}B$(i.e., the distributions of $Y_m$ converge to to the distribution of BM $B$ weakly on $C[0,1]$), get \begin{gather*} \lim_{m\to\infty}w(Y_m,\delta)\stackrel{d}{=} w(B,\delta).\\ \lim_{\delta\to0}\varlimsup_{m\to\infty} \mathsf{P}(w(Y_m,\delta)>\epsilon)=0, \quad \forall \epsilon>0.\tag{5} \end{gather*} Next, the following fact will be proved, \begin{equation*} \lim_{m\to\infty}\mathsf{P}(\sup_{0\le t\le 1}|U_m(t)|>\epsilon)=0, \quad \forall \epsilon>0. \tag{6} \end{equation*} Denote \begin{equation*} A_m(\delta)=\Big\{\omega: \sup_{0\le t\le 1}\Big| \frac{\alpha(mt)}{m} \Big| \ge \delta \Big\}, \end{equation*} then from (1) \begin{gather*} \lim_{m\to\infty}\mathsf{P}(A_m(\delta))=0, \qquad \forall \delta>0,\\ \lim_{m\to\infty}\mathsf{P}([\sup_t|U_m(t)|]I_{A_m(\delta)}>\epsilon)=0, \qquad \forall \delta, \epsilon>0. \tag{7} \end{gather*} On the other hand, \begin{gather*} |U_m(t)| I_{A_m^c(\delta)}\le w(Y_m,\delta), \qquad \forall t\in[0,1],\\ [\sup_{0\le t\le 1}|U_m(t)|]I_{A_m^c(\delta)}\le w(Y_m,\delta). \tag{8} \end{gather*} From (8),(5), get \begin{equation*} \lim_{\delta\to0}\varlimsup_{m\to\infty} \mathsf{P}([\sup_{0\le t\le 1}|U_m(t)|]I_{A_m^c(\delta)}>\epsilon)=0, \quad \forall \epsilon>0.\tag{9} \end{equation*} Hence, from (7) and (9), (6) is true. Furthermore, $Z_m\stackrel{d}{\longrightarrow}B$ comes from $Y_m\stackrel{d}{\longrightarrow}B$ and (4),(6).